Question

Let $y=y\left(x\right)$ be the solution of the differential equation $dy=e^{\alpha x+y}dx;$ $\alpha \in N.$ If $y\left({\log }_{e}2\right)={\log }_{e}2$ and $y\left(0\right)={\log }_e\left(\frac{1}{2}\right)$, then the value of $\alpha$ is equal to

Answer 1

$e^{-y}dy=e^{\alpha x}dx$
$\Rightarrow -e^{-y}=\frac{1}{\alpha }{e}^{\alpha x}+c$
Put $x=y={\log }_{e}2$ and $x=0,y=-{\log }_{e}2$
$-\frac{1}{2}=\frac{2^{\alpha }}{\alpha }+c$ and $-2=\frac{1}{\alpha }+c$
$\Rightarrow \alpha =2$